How do you determine if the improper integral converges or diverges #int (1/(3x)-6) dx# from negative infinity to 0?

1 Answer
Mar 25, 2018

#int_-oo^0(1/(3x)-6)dxrArr# diverges to #-oo#

Explanation:

To evaluate a two-sided improper integral, split it into two integrals and express each as a limit like so:

#int_-oo^0(1/(3x)-6)dx#

#=lim_(s->-oo)int_s^-1(1/(3x)-6)dx+lim_(t->0)int_-1^t(1/(3x)-6)dx#

Now compute the integrals and evaluate the limits:

#rArrlim_(s->-oo)[lnabs(3x)/3-6x]_s^-1#

#rArrlim_(s->-oo)[(ln3/3+6)-(lnabs(3s)/3-6s)]#

#rArr[(ln3/3+6)-(oo+oo)]->-oo#

Therefore the integral diverges as it approaches #-oo#.

Technically, we could stop here. If one integral diverges, the whole expression will diverge. But let's evaluate the second limit just for good measure:

#rArrlim_(t->0)[lnabs(3x)/3-6x]_-1^t#

#rArrlim_(t->0)[(lnabs(3t)/3-6t)-(ln3/3+6)]#

#rArr[(-oo-0)-(ln3/3+6)]->-oo#

And again we see the integral also diverges as it approaches 0.